Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

1. Intermediate Maths Challenge Problem (UKMT)

4 marks

1.1 Four congruent isosceles trapeziums are placed so that their longer parallel sides form the diagonals of a square PQRS as shown.

The point X divides PQ in the ratio 3:1.

What fraction of the square is shaded?

56

38

716

512

12

Show Hint (–1 mark)

–1 mark

Since X divides PQ in the ratio 3:1, it is convenient to choose units so that the side of the square has length 4, and hence XQ has length 1. Then the total area of the square is 42=16.

Show Hint (–1 mark)

–1 mark

If we add the dotted lines to the diagram, as shown on the right, we see that the unshaded part of the square is made up of 4 isosceles right-angled triangles in which the shorter sides have length 2, and 4 isosceles right-angled triangles in which the shorter sides have length 1.

Since X divides PQ in the ratio 3:1, it is convenient to choose units so that the side of the square has length 4, and hence XQ has length 1. Then the total area of the square is 42=16.

If we add the dotted lines to the diagram, as shown on the right, we see that the unshaded part of the square is made up of 4 isosceles right-angled triangles in which the shorter sides have length 2, and 4 isosceles right-angled triangles in which the shorter sides have length 1. Each of the larger of these triangles has area 1222=2, and
each of the smaller triangles has area 1212=1. Therefore the total area of the square that is not shaded is 4×2+4×12=10.

Therefore the area that is shaded is 16−10=6. Hence the fraction of the square that is shaded is 616=38.

2. Intermediate Maths Challenge Problem (UKMT)

3 marks

2.1 Which of the following has the greatest value?

1173

533

743

953

323

Show Hint (–1 mark)

–1 mark

The fact that each bracket is cubed is a distraction. If A is bigger than B, and if A is bigger than 1, then A3 is bigger than B3.

Show Hint (–1 mark)

–1 mark

There are several approaches, but one way is to make every fraction have the same denominator, e.g., 420.

If x and y are positive numbers, then x<y if and only if x3<y3. So to determine which of the given fractions when cubed results in the largest number, we need only decide which of the fractions is the largest.

To do this we use the fact that, where p, q, r and s are positive integers, we have pq<rs⇔ps<rq, and so we can compare the size of the fractions pq and rs by comparing the size of the integers ps and qr.

Since 11×3<5×7, 117<53, and since 5×4<7×3, 53<74.

Also 3×5<9×2 and so 32<95. Hence the largest fraction is either 74 or 95. Since 7×5<9×4, 74<95, and therefore 95 is the largest of the given fractions.

We deduce that 953 is the largest of the given cubes.

3. QI Mathematics

If you watch QI on BBC2, then you will know that it is full of quite interesting things, including maths. This video is a compilation of some of the interesting maths that has been featured in the programme. Watch carefully, and then answer the questions below.

1 mark

3.1 Let’s look at McNugget numbers. McNuggets come in boxes of 6, 9 and 20. The biggest number of nuggets that cannot be made from combinations of 6, 9 and 20 is 43 nuggets, as mentioned in the video. But can you buy 37 nuggets?

Yes

No

1 mark

3.2 Can you buy 38 nuggets?

Yes

No

38 = 20 + 9 x 2 or 20 + 6 x 3.

1 mark

3.3 Can you buy 39 nuggets?

Yes

No

39 = 6 x 5 + 9.

1 mark

3.4 Can you buy 40 nuggets?

Yes

No

40 = 20 + 20.

1 mark

3.5 Can you buy 41 nuggets?

Yes

No

41 = 20 + 9 + 2 x 6.

1 mark

3.6 Can you buy 42 nuggets?

Yes

No

42 = 6 x 7.

1 mark

3.7 What is the surname of the philosopher mentioned in the QI clip who proved that 1 + 1 = 2? It contains 7 letters.

Correct Solution: RUSSELL

2 marks

3.8 Pick a 3-digit number (with different digits), then create a new 3-digit number by reversing the digits. Take the smaller 3-digit number from the bigger 3-digit number to create a third 3-digit number (which we’ll call the “difference number”). Now reverse the digits of the “difference number” and add the result to the “difference number” – what is the result?

(The result should be same whatever you decide to chose for your first 3-digit number. Think about why this might be.)

Correct Solution: 1089

4. Intermediate Maths Challenge Problem (UKMT)

5 marks

4.1. In 1984 the engineer and prolific prime-finder Harvey Dubner found the biggest known prime each of whose digits is either a one or a zero.

The prime can be expressed as 10641×10640−19+1.

How many digits does this prime have?

640

641

1280

1281

640 × 641

Show Hint (–1 mark)

–1 mark

The answer is odd.

The number 10^640 − 1 consists of a string of 640 nines when written out in standard form. So the
number 10641×10640–1 consists of a string of 640 nines followed by 641 zeros. Therefore 10641×10640–19 consists of a string of 640 ones followed by 641 zeros. Adding 1 just changes the final digit from zero to a one. So the prime is: